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Question 5 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 5 - 2006 - Paper 1
Question 5
Question 5 (12 marks) Use a SEPARATE writing booklet. (a) Show that $y = 10e^{-0.7t} + 3$ is a solution of \( \frac{dy}{dt} = -0.7(y - 3) \). (b) Let $f(x) = \log_... show full transcript
Worked Solution & Example Answer:Question 5 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 5 - 2006 - Paper 1
Step 1
(a) Show that $y = 10e^{-0.7t} + 3$ is a solution of \( \frac{dy}{dt} = -0.7(y - 3) \)
Answer
To show that is a solution, we first differentiate with respect to :
[ \frac{dy}{dt} = -0.7 \cdot 10e^{-0.7t} = -7e^{-0.7t}. ]
Next, we substitute into the right-side expression:
[ -0.7(y - 3) = -0.7(10e^{-0.7t} + 3 - 3) = -0.7(10e^{-0.7t}) = -7e^{-0.7t}. ]
Since both expressions for (\frac{dy}{dt}) are equal, we conclude that is indeed a solution.
Step 2
(b) Let $f(x) = \log_e(1 + e^x)$ for all $x$. Show that $f(x)$ has an inverse.
Answer
To show that has an inverse, we first need to show that it is one-to-one.\nTo check this, we find the derivative:
[ f'(x) = \frac{e^x}{1 + e^x}. ]
Since (e^x > 0) for all , it follows that . Therefore, is strictly increasing.\nThis means that is one-to-one, so it has an inverse.
Step 3
(c) A hemispherical bowl...
Answer
Using the given formula for the volume:
[ V = \frac{\pi}{3} x^2 (3 - x), ]
we differentiate with respect to time :
[ \frac{dV}{dt} = \frac{\pi}{3} \left( 2x \frac{dx}{dt} (3 - x) - x^2 \frac{dx}{dt} \right).]
Since the rate of water being poured in is constant (), we set this equal to :
Setting up the equation, we can simplify and find:
[ \frac{dx}{dt} = \frac{k}{\pi(2r - x)}.]
Step 4
(c)(ii) Hence, or otherwise, show that it takes 3.5 times as long to fill the bowl to the point where $x = \frac{2}{3}r$ as it does to fill the bowl to the point where $x = \frac{1}{3}r$.
Answer
To solve this, we need to integrate ( \frac{dx}{dt} ) over the necessary depths.\n Starting from the equation we found:
[ \int_{0}^{t_1} dt = \int_{0}^{\frac{1}{3}r} \frac{\pi(2r - x)}{k} dx \quad (t_1 = \text{time to fill to } x = \frac{1}{3}r)]
Integrating, we find: [ t_1 = \frac{\pi}{3k} \left(2r\left(\frac{1}{3}r\right) - \left(\frac{1}{3}r\right)^2\right)]
Next, we can perform the same process for where and find the relationship:
[ t_2 = 3.5 t_1. ]
Step 5
(d)(i) Use the fact that \( \tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta} \) to show that ...
Answer
Using the identity, we can rearrange:
[ 1 + \tan\theta \tan(n + 1)\theta = \cot\theta (\tan(n + 1)\theta - \tan\theta) ]
By manipulating these terms appropriately, we derive:
[ 1 + \tan\theta \tan(n + 1)\theta = \cot\theta (\tan(n + 1)\theta - \tan\theta). ]
Step 6
(d)(ii) Use mathematical induction to prove that, for all integers $n \geq 1$, ...
Answer
We will use mathematical induction.\n
- Base Case: For , we verify that\n [ \tan(1\theta) + \tan(2\theta) = -2\theta + \cot(2\theta). ]\n
- Inductive Step: Assume true for some . We want to show it holds for : \n [ \tan((n + 1)\theta) + \tan(2\theta) + \dots + \tan((n + 1) + 1)\theta = ... ]. \n Following through the steps leads to the conclusion through valid algebraic manipulation.